solve for ( cos &#x2061;<!-- ⁡ --> x ) ( cos &#x2061;<!-- ⁡ --> 2 x

pachaquis3s

pachaquis3s

Answered question

2022-06-26

solve for ( cos x ) ( cos 2 x ) ( cos 3 x ) = 1 4

Answer & Explanation

pheniankang

pheniankang

Beginner2022-06-27Added 22 answers

cos x cos 2 x cos 3 x = 1 / 4
( 2 cos x cos 3 x ) ( 2 cos 2 x ) = 1
( cos 4 x + cos 2 x ) ( 2 cos 2 x ) = 1
2 cos 4 x cos 2 x + ( 2 cos 2 2 x 1 ) = 0
2 cos 4 x cos 2 x + cos 4 x = 0
cos 4 x   ( 2 cos 2 x + 1 ) = 0
Either,   cos 4 x = 0 4 x = ( 2 n + 1 ) π 2 x = ( 2 n + 1 ) π 8   , for any integer n .
Or,   2 cos 2 x + 1 = 0 2 cos 2 x = 1 cos 2 x = 1 2 cos 2 x = cos ( π π 3 )  
  cos 2 x = cos ( 2 π 3 ) 2 x = 2 n π ± 2 π 3 x = n π ± π 3   , for any integer n .

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?